Soling a boundary value problem

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Sharon 2024-2-2

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评论： Sharon 2024-2-3

I am trying to solve the following elliptic equation

on the interval [0,1] with Neumann Boundary condition,

I want to get a positive solution φ with the maximum of φ is 1.

The following is the code I am using

h=0.001;

L=1;

M=L/h;

x=linspace(0,L,M+1);

gamma = @(x)(1+0.1*sin(2*pi*x)).*(1+0.1*sin(2*pi*x));

beta=@(x)(1+0.1*sin(2*pi*x));

%v0 = @(x)(1+0.5*cos(pi*x));

%v0x=@(x)-0.5*pi*sin(pi*x);

bvpfcn = @(x,v)[v(2);(gamma(x)-(beta(x)).*v(1)];

bcfcn = @(va,vb)[va(2);vb(2)];

guess = @(x)[1;0];

%guess = @(x)[v0(x);v0x(x)];

solinit = bvpinit(x,guess);

sol = bvp4c(bvpfcn, bcfcn, solinit);

phi1=sol.y(1,:);

M=max(phi1);

phi=phi1/M;

plot(sol.x,phi)

I have difficulity in selecting the initial function. I tried [1;0] and [v0(x);v0x(x)] which is defined above, I got different solutions. I cannot figure out the problem. Any help would be appreciated!

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回答（1 个）

Torsten 2024-2-3

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编辑：Torsten 2024-2-3

Prove the following:

If u is a solution of your differential equation, so is c*u for every c in IR.

So it does not surprise that bvp4c gives different solutions dependent on your initial guess: the solution is not unique.

If you normalize the solution as you did after solving, it seems it becomes unique. You can see this if you solve your equation for your two initial guess functions and plot the solutions in one graph.